Problem: What is the slope of the secant line that intersects the graph of $g(x)=\dfrac{x}{x^2+1}$ at $x=2$ and $x=3$ ?
Answer: The secant line will pass through points $(2,g(2))$ and $(3,g(3))$. $\begin{aligned} \text{slope}&=\dfrac{\text{change in }y}{\text{change in }x} \\\\ &=\dfrac{g(3)-g(2)}{3-2} \end{aligned}$ ${1}$ ${2}$ ${3}$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{\llap{-}1}{5}$ $\frac{\llap{-}2}{5}$ $y$ $x$ $(2,g(2))$ $(3,g(3))$ secant line We will need to know the values of $g(2)$ and $g(3)$ to find the slope. $\begin{aligned} g(2)&=\dfrac{2}{2^2+1} \\\\ &=\dfrac{2}{5} \\\\ g(3)&=\dfrac{3}{3^2+1} \\\\ &=\dfrac{3}{10} \end{aligned}$ Now we can find the slope: $\begin{aligned} \dfrac{g(3)-g(2)}{3-2}&=\dfrac{\dfrac{3}{10}-\dfrac{2}{5}}{1} \\\\ &=-\dfrac{1}{10} \end{aligned}$ The slope of the secant line that intersects the graph of $g(x)=\dfrac{x}{x^2+1}$ at $x=2$ and $x=3$ is $-\dfrac{1}{10}$. Notice that the slope of the secant line is calculated just like the average rate of change over the interval.